# How do you find the integral ln(x) x^(3/2) dx?

Jun 18, 2015

To do integration by parts:

$u v = \int v \mathrm{du}$

I would pick a $u$ that gives me an easy time differentiating, and a $\mathrm{dv}$ that I would easily be able to integrate multiple times if necessary. If the same integral comes back, or variables aren't going away, either it's cyclic or you should switch your $u$ and $\mathrm{dv}$.

Let:
$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dv} = {x}^{\frac{3}{2}} \mathrm{dx}$
$v = \frac{2}{5} {x}^{\frac{5}{2}}$

$\implies \frac{2}{5} {x}^{\frac{5}{2}} \ln x - \int \frac{2}{5} \left({x}^{\frac{5}{2}} / x\right) \mathrm{dx}$

$= \frac{2}{5} {x}^{\frac{5}{2}} \ln x - \int \frac{2}{5} {x}^{\frac{3}{2}} \mathrm{dx}$
$= \frac{2}{5} {x}^{\frac{5}{2}} \ln x - \frac{4}{25} {x}^{\frac{5}{2}} + C$
$= \frac{10}{25} {x}^{\frac{5}{2}} \ln x - \frac{4}{25} {x}^{\frac{5}{2}} + C$
$= {x}^{\frac{5}{2}} \left[\frac{10}{25} \ln x - \frac{4}{25}\right] + C$
$= \frac{2}{25} {x}^{\frac{5}{2}} \left[5 \ln x - 2\right] + C$